Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <class TH>
void _dbg(const char *sdbg, TH h) {
  cerr << sdbg << "=" << h << "\n";
}
template <class TH, class... TA>
void _dbg(const char *sdbg, TH h, TA... a) {
  while (*sdbg != ',') cerr << *sdbg++;
  cerr << "=" << h << ",";
  _dbg(sdbg + 1, a...);
}
const int MX = 107;
int t[MX][MX];
int minCol[MX], minRow[MX];
int segR[MX], segC[MX];
int n, m;
vector<pair<int, int> > odp;
int sum;
int fun() {
  for (int i = 1; i < MX; ++i) {
    minRow[i] = INT_MAX;
    minCol[i] = INT_MAX;
    segR[i] = 0;
    segC[i] = 0;
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= m; ++j) {
      minRow[i] = min(minRow[i], t[i][j]);
      minCol[j] = min(minCol[j], t[i][j]);
    }
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= m; ++j) {
      segC[j] = max(segC[j], t[i][j] - minRow[i]);
      segR[i] = max(segR[i], t[i][j] - minCol[j]);
    }
  }
  for (int i = 1; i <= n; ++i) {
    if (segR[i] > minRow[i]) return -1;
  }
  for (int i = 1; i <= m; ++i) {
    if (segC[i] > minCol[i]) return -1;
  }
  if (m >= n) {
    for (int i = 1; i <= n; ++i) {
      if (minRow[i] > 0) {
        odp.push_back({i, minRow[i]});
        sum += minRow[i];
        for (int j = 1; j <= m; ++j) t[i][j] -= minRow[i];
        return 1;
      }
    }
    for (int i = 1; i <= m; ++i) {
      if (minCol[i] > 0) {
        odp.push_back({-i, minCol[i]});
        sum += minCol[i];
        for (int j = 1; j <= n; ++j) t[j][i] -= minCol[i];
        return 1;
      }
    }
  } else {
    for (int i = 1; i <= m; ++i) {
      if (minCol[i] > 0) {
        odp.push_back({-i, minCol[i]});
        sum += minCol[i];
        for (int j = 1; j <= n; ++j) t[j][i] -= minCol[i];
        return 1;
      }
    }
    for (int i = 1; i <= n; ++i) {
      if (minRow[i] > 0) {
        odp.push_back({i, minRow[i]});
        sum += minRow[i];
        for (int j = 1; j <= m; ++j) t[i][j] -= minRow[i];
        return 1;
      }
    }
  }
  return -1;
}
bool check() {
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= m; ++j)
      if (t[i][j] != 0) return 0;
  return 1;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= m; ++j) cin >> t[i][j];
  int res = 1;
  while (res > 0) {
    res = fun();
  }
  if (!check()) return cout << -1, 0;
  cout << sum << '\n';
  for (pair<int, int> p : odp) {
    int N = p.first, ile = p.second;
    for (int i = 0; i < ile; ++i)
      cout << (N > 0 ? "row" : "col") << " " << abs(N) << '\n';
  }
}